Determine magnitude of force acting on block on inclined plane

[JJBladester]

Homework Statement

A 20-kg package is at rest on an incline when a force P is applied to it. Determine the magnitude of P if 10s is required for the package to travel 5m up the incline. The static and kinetic coefficients of friction between the package and the incline are both equal to 0.3.

Answer: 419N to start and 301 during sliding

Homework Equations

m=20kg
g=9.8m/s²
t=10s
x_i = 0ft
x_f = 5ft
μ_s = μ_k = 0.3
Frictional force = F_f = μN

The Attempt at a Solution

First, I drew a free-body diagram including tilted coordinate axes:

Then, I went about finding the sum of the forces equations in the x- and y-directions.

\sum F_{x}=-F_{f}-mgsin(20)+Pcos(30)=ma

\sum F_{y}=N-mgcos(20)-Psin(30)=0

From the second equation,
N=mgcos(20)+Psin(30)

Plugging this value for N into F_f = μN in the first equation yields:

P=\frac{m\left(a+\mu gcos(20)+gsin(20)\right)}{cos(30)-\mu sin(30)}

If I know the acceleration, I can find the magnitude of P. There has to be a way to find acceleration from the initial conditions, but I'm at a loss. Also, once static friction is overcome, I'm sure the sum of the forces equations will be different but I'm not sure how.

 

[Thefox14]

Well since the force seems to be constant, you could apply the kinematic equation to solve for the acceleration:
x = x_0 + v_0t + (1/2)at^2

 

[JJBladester]

Well since the force seems to be constant, you could apply the kinematic equation to solve for the acceleration:
x = x_0 + v_0t + (1/2)at^2

From the \sum F_{x} equation, along with F_{f}=\mu N:

a=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)

x_{f}=x_{i}+v_{i}t+\frac{1}{2}at^{2}

5=\frac{1}{2}\left (-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30) \right )\left (10^{2} \right )

\frac{1}{10}=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)

\frac{1}{10}+\mu gcos(20)+gsin(20)=P\left (cos(30)-\frac{\mu sin(30)}{m} \right )

P=\frac{\frac{1}{10}+\mu gcos(20)+gsin(20)}{cos(30)-\frac{\mu sin(30)}{m}}=\frac{\frac{1}{10}+(0.3)(9.8)cos(20)+(9.8)sin(20)}{cos(30)-\frac{0.3sin(30)}{20}}\approx 7.24N \neq \: Book's\: answer\: of\: 419N

Did I do the math wrong? Where do I go from here?